Question

Determining Ionic Strength

Unlike strong acids and bases that ionize completely in solution, weak acids or bases partially ionize. The tendency to ionize (i.e., the ionic strength) of a weak acid or base can be quantified in several ways including

Ka or Kb,

pKa or pKb,

and percent ionization.

Part A

Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows:

C5H5N+H2O?C5H5NH++OH?

The pKb of pyridine is 8.75. What is the pH of a 0.150M solution of pyridine?

pH =

Part B

Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid ionizes in water:

C6H5COOH?C6H5COO?+H+

The pKa of this reaction is 4.2. In a 0.51Msolution of benzoic acid, what percentage of the molecules are ionized?

percent ionized =

Answer 1

Part-A: Ionization reaction of pyridine in water is

         C5H5N+H2O ---> C5H5NH⁺ + OH^-

Init:    0.150M 0M             0M

Eqm: (0.150-x)M xM             xM

Given pKb of pyridine is 8.75

=> - logKb = 8.75

=> Kb = antilog(-8.75) = 1.78*10^-9

Now Kb can also be calculated as

Kb = [C5H5NH⁺]*[OH^-] / [C5H5N] = x*x / (0.150-x) = x² /(0.150-x)    

= >1.78*10^-9 = x² /(0.150-x)    

Since x << 0.150, (0.150-x) is nearly equals to 0.150

=> x = underroot(0.150*1.78*10^-9) = 1.63*10^-5 M

=> x = [OH^-] = 1.63*10^-5 M

=> pOH = - log[OH^-] = -log(1.63*10^-5) = 4.79

=> pH = 14 – pOH = 14 – 4.79 = 9.21 (answer)

Part-B: Let the degree of ionization of benzoic acid be $\alpha$

Given [C6H5COOH], C = 0.51M

C6H5COOH --> C6H5COO^- + H⁺

Init: 0.51M 0M 0M

eqm: 0.51(1 - $\alpha$ )M 0.51 $\alpha$ M 0.51 $\alpha$ M

Given pKa = 4.2

=> - logKa = 4.2

=>Ka = antilog(-4.2) = 6.3*10^-5

Now Ka can also be calculated as

Ka = [C6H5COO^- ]*[H⁺] / [C6H5COOH] = ( 0.51 $\alpha$ M)*(0.51 $\alpha$ M) / 0.51(1 - $\alpha$ )M

=>  6.3*10^-5 = ( 0.51 $\alpha$ M)*(0.51 $\alpha$ M) / 0.51(1 - $\alpha$ )M

Since $\alpha$ <<1 , (1 - $\alpha$ ) is nearly equals to 1. Hence

  6.3*10^-5 = 0.51 $\alpha$ ²

=> $\alpha$ = 1.11x10^-2

Hence percentage ionisation = $\alpha$ *100 = 1.11 % (answer)